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feat: add solutions to lc problems: No.2833~2836 (#1519)
* No.2833.Furthest Point From Origin * No.2834.Find the Minimum Possible Sum of a Beautiful Array * No.2835.Minimum Operations to Form Subsequence With Target Sum * No.2836.Maximize Value of Function in a Ball Passing Game
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‎solution/1400-1499/1483.Kth Ancestor of a Tree Node/README.md

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时间复杂度方面，初始化为 $O(n \times \log n)$，查询为 $O(\log n)$。空间复杂度 $O(n \times \log n)$。其中 $n$ 为树的节点数。
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相似题目：
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- [2836. 在传球游戏中最大化函数值](/solution/2800-2899/2836.Maximize%20Value%20of%20Function%20in%20a%20Ball%20Passing%20Game/README.md)
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<!-- tabs:start -->
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### **Python3**

‎solution/2800-2899/2833.Furthest Point From Origin/README.md

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# [2833. 距离原点最远的点](https://leetcode.cn/problems/furthest-point-from-origin)
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[English Version](/solution/2800-2899/2833.Furthest%20Point%20From%20Origin/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个长度为 <code>n</code> 的字符串 <code>moves</code> ，该字符串仅由字符 <code>'L'</code>、<code>'R'</code> 和 <code>'_'</code> 组成。字符串表示你在一条原点为 <code>0</code> 的数轴上的若干次移动。</p>
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<p>你的初始位置就在原点（<code>0</code>），第 <code>i</code> 次移动过程中，你可以根据对应字符选择移动方向：</p>
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<ul>
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<li>如果 <code>moves[i] = 'L'</code> 或 <code>moves[i] = '_'</code> ，可以选择向左移动一个单位距离</li>
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<li>如果 <code>moves[i] = 'R'</code> 或 <code>moves[i] = '_'</code> ，可以选择向右移动一个单位距离</li>
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</ul>
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<p>移动 <code>n</code> 次之后，请你找出可以到达的距离原点 <strong>最远</strong> 的点，并返回 <strong>从原点到这一点的距离</strong> 。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>moves = "L_RL__R"
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<strong>输出：</strong>3
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<strong>解释：</strong>可以到达的距离原点 0 最远的点是 -3 ，移动的序列为 "LLRLLLR" 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>moves = "_R__LL_"
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<strong>输出：</strong>5
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<strong>解释：</strong>可以到达的距离原点 0 最远的点是 -5 ，移动的序列为 "LRLLLLL" 。
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre>
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<strong>输入：</strong>moves = "_______"
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<strong>输出：</strong>7
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<strong>解释：</strong>可以到达的距离原点 0 最远的点是 7 ，移动的序列为 "RRRRRRR" 。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= moves.length == n &lt;= 50</code></li>
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<li><code>moves</code> 仅由字符 <code>'L'</code>、<code>'R'</code> 和 <code>'_'</code> 组成</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心**
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遇到字符 `'_'` 时，我们可以选择向左或向右移动，而题目需要我们求出离原点最远的点，因此，我们可以先进行一次遍历，贪心地把所有的 `'_'` 都移到左边，求出此时离原点最远的点，再进行一次遍历，贪心地把所有的 `'_'` 都移到右边，求出此时离原点最远的点，最后取两次遍历中的最大值即可。
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进一步地，我们只需要统计出字符串中 `'L'`、`'R'` 的个数之差，再加上 `'_'` 的个数即可。
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时间复杂度 $O(n)$，其中 $n$ 为字符串的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def furthestDistanceFromOrigin(self, moves: str) -> int:
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return abs(moves.count("L") - moves.count("R")) + moves.count("_")
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int furthestDistanceFromOrigin(String moves) {
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return Math.abs(count(moves, 'L') - count(moves, 'R')) + count(moves, '_');
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}
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private int count(String s, char c) {
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int cnt = 0;
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for (int i = 0; i < s.length(); ++i) {
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if (s.charAt(i) == c) {
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++cnt;
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}
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}
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return cnt;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int furthestDistanceFromOrigin(string moves) {
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auto cnt = [&](char c) {
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return count(moves.begin(), moves.end(), c);
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};
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return abs(cnt('L') - cnt('R')) + cnt('_');
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}
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};
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```
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### **Go**
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```go
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func furthestDistanceFromOrigin(moves string) int {
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count := func(c string) int { return strings.Count(moves, c) }
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return abs(count("L")-count("R")) + count("_")
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **TypeScript**
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```ts
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function furthestDistanceFromOrigin(moves: string): number {
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const count = (c: string) => moves.split('').filter(x => x === c).length;
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return Math.abs(count('L') - count('R')) + count('_');
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2800-2899/2833.Furthest Point From Origin/README_EN.md

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# [2833. Furthest Point From Origin](https://leetcode.com/problems/furthest-point-from-origin)
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[中文文档](/solution/2800-2899/2833.Furthest%20Point%20From%20Origin/README.md)
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## Description
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<p>You are given a string <code>moves</code> of length <code>n</code> consisting only of characters <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and <code>&#39;_&#39;</code>. The string represents your movement on a number line starting from the origin <code>0</code>.</p>
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<p>In the <code>i<sup>th</sup></code> move, you can choose one of the following directions:</p>
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<ul>
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<li>move to the left if <code>moves[i] = &#39;L&#39;</code> or <code>moves[i] = &#39;_&#39;</code></li>
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<li>move to the right if <code>moves[i] = &#39;R&#39;</code> or <code>moves[i] = &#39;_&#39;</code></li>
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</ul>
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<p>Return <em>the <strong>distance from the origin</strong> of the <strong>furthest</strong> point you can get to after </em><code>n</code><em> moves</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> moves = &quot;L_RL__R&quot;
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> The furthest point we can reach from the origin 0 is point -3 through the following sequence of moves &quot;LLRLLLR&quot;.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> moves = &quot;_R__LL_&quot;
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> The furthest point we can reach from the origin 0 is point -5 through the following sequence of moves &quot;LRLLLLL&quot;.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> moves = &quot;_______&quot;
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<strong>Output:</strong> 7
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<strong>Explanation:</strong> The furthest point we can reach from the origin 0 is point 7 through the following sequence of moves &quot;RRRRRRR&quot;.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= moves.length == n &lt;= 50</code></li>
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<li><code>moves</code> consists only of characters <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code> and <code>&#39;_&#39;</code>.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def furthestDistanceFromOrigin(self, moves: str) -> int:
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return abs(moves.count("L") - moves.count("R")) + moves.count("_")
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```
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### **Java**
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```java
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class Solution {
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public int furthestDistanceFromOrigin(String moves) {
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return Math.abs(count(moves, 'L') - count(moves, 'R')) + count(moves, '_');
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}
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private int count(String s, char c) {
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int cnt = 0;
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for (int i = 0; i < s.length(); ++i) {
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if (s.charAt(i) == c) {
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++cnt;
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}
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}
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return cnt;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int furthestDistanceFromOrigin(string moves) {
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auto cnt = [&](char c) {
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return count(moves.begin(), moves.end(), c);
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};
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return abs(cnt('L') - cnt('R')) + cnt('_');
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}
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};
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```
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### **Go**
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```go
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func furthestDistanceFromOrigin(moves string) int {
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count := func(c string) int { return strings.Count(moves, c) }
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return abs(count("L")-count("R")) + count("_")
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **TypeScript**
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```ts
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function furthestDistanceFromOrigin(moves: string): number {
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const count = (c: string) => moves.split('').filter(x => x === c).length;
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return Math.abs(count('L') - count('R')) + count('_');
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2800-2899/2833.Furthest Point From Origin/Solution.cpp

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class Solution {
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public:
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int furthestDistanceFromOrigin(string moves) {
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auto cnt = [&](char c) {
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return count(moves.begin(), moves.end(), c);
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};
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return abs(cnt('L') - cnt('R')) + cnt('_');
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}
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};

‎solution/2800-2899/2833.Furthest Point From Origin/Solution.go

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func furthestDistanceFromOrigin(moves string) int {
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count := func(c string) int { return strings.Count(moves, c) }
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return abs(count("L")-count("R")) + count("_")
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}

‎solution/2800-2899/2833.Furthest Point From Origin/Solution.java

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class Solution {
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public int furthestDistanceFromOrigin(String moves) {
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return Math.abs(count(moves, 'L') - count(moves, 'R')) + count(moves, '_');
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}
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private int count(String s, char c) {
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int cnt = 0;
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for (int i = 0; i < s.length(); ++i) {
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if (s.charAt(i) == c) {
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++cnt;
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}
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}
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return cnt;
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}
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}

‎solution/2800-2899/2833.Furthest Point From Origin/Solution.py

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class Solution:
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def furthestDistanceFromOrigin(self, moves: str) -> int:
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return abs(moves.count("L") - moves.count("R")) + moves.count("_")

‎solution/2800-2899/2833.Furthest Point From Origin/Solution.ts

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function furthestDistanceFromOrigin(moves: string): number {
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const count = (c: string) => moves.split('').filter(x => x === c).length;
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return Math.abs(count('L') - count('R')) + count('_');
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}

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